What type of reaction leads to the formation of a precipitate?

Sep 13, 2017

Typically, we examine a $\text{solubility equilibrium.....}$

Explanation:

Typically, we examine a solubility equilibrium, where concentrations of (metal) cation, and (non-metal) anion occur such that their product exceeds a given constant, i.e. ${K}_{\text{sp}}$.

And we can take one hallowed example......

$A {g}^{+} + C {l}^{-} r i g h t \le f t h a r p \infty n s A g C l \left(s\right) \downarrow$

For standard conditions, a ${K}_{\text{sp}}$ value can be measured such that.....

${K}_{\text{sp}} = \left[A {g}^{+}\right] \left[C {l}^{-}\right] = 1.77 \times {10}^{-} 10$

And if the product $\left[A {g}^{+}\right] \left[C {l}^{-}\right] \ge 1.77 \times {10}^{-} 10$, then precipitation WILL occur.

See here and links for further detail....

For standard reactions in the inorganic laboratory, we use general rules to assess aqueous solubility: all the salts of the alkali metals are soluble; all the nitrates are soluble; all halides are soluble except for $A g X$, $H {g}_{2} {X}_{2}$, $P b {X}_{2}$.......; all sulfides and phosphates and hydroxides are insoluble......; all hydroxides are insoluble.....etc. etc.